Problem: A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".

You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.

Determine if it is possible to obtain an RBS after these replacements.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

Output

For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}. 

You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

Input


5
()
(?)
(??)
??()
)?(?


Output


YES
NO
YES
YES
NO

Note

In the first test case, the sequence is already an RBS.

In the third test case, you can obtain an RBS as follows: ()() or (()).

In the fourth test case, you can obtain an RBS as follows: ()().

Solution (in PYTHON):

Solution:
def check(s):
    stack=[]
    for i in s:
        if i=="(" or len(stack)==0:
            stack+=[i]
        elif i=="?":
            return 0
        else:
            stack.pop()
    if len(stack)==0:return 1
    else:return 0
def pre(s):
    a=0
    for i in range(len(s)):
        if s[i]==')':
            a+=1
        if a>(i+1)/2:
            return 0
    return 1
def solve(s):
    l=''
    r=''
    a,b=s.count('('),s.count(")")
    half=len(s)/2
    if a>half or b>half:
        return 0
    if len(s)%2==1:
        return 0
    if pre(s)==0:
        return 0
    for i in s:
        if i=="?":
            l+='('
            r+=')'
        else:
            l+=i
            r+=i
    if check(s)==1 or check(l)==1 or check(r)==1:
        return  1
    else:
        l=''
        a,b=0,0
        for i in s:
            if i=="(":
                a+=1
            elif i==")":
                b+=1
            else:
                if a==b or a<b:
                    l+='('
                else:
                    l+=")"
        return check(l)
for _ in range(input()):
    s=raw_input()
    if solve(s)==1:
        print "YES"
    else:
        print "NO"